Question

how to factorise a^2+c^2-b^2+2ac using identities

Answer 1

ITS AN EXAMPLE:

We want to factorise  c2−a2−b2−2ab .

Let’s manipulate it a bit and see what we get:

c2−a2−b2−2ab  

=c2−(a2+2ab+b2)  

Now we can see that

a2+2ab+b2  is a perfect square, an algebraic identity we can use, and can be factorised as thus:

a2+2ab+b2  

=a2+ab+ab+b2  

=a(a+b)+b(a+b)  

=(a+b)(a+b)  

=(a+b)2  

Now we substitute this back into our expression:

c2−(a+b)2  

(a+b)(a−b)=a2−ab+ab−b2  

=a2−b2  

So, we can see that  a2−b2=(a+b)(a−b) . Now let’s use this identity to simplify our expression:

c2−(a+b)2  

=(c+a+b)(c−a−b)  

Therefore, the factorised form of the original expression is  (c+a+b)(c−a−b) .

Answer 2

a^2–2ab+b^2 - c^2

((a-b)(a-b))-c^2

(a-b)^2-c^2

Set x =a-b and y=c. The formula becomes

x^2-y^2

factoring this polynomial, we get

(x+y)(x-y)

Substituting back, we get:

(a-b+c)(a-b-c)

Let’s multiply it out to check:

A^2 -ab ac

- ab B^2 -bc

-ac bc -c^2

Boom. QED. LOL.

Break it down into parts:

a²+b²-2ab=(a-b)(a-b)

a²+b²-c²-2ab=(a-b)²-c²

Then, factor in c²:

(a-b)²-c²=((a-b)+c)((a-b)-c)

=(a-b+c)(a-b+c).....

HOPE IT HELPS UH